Question

which term of the AP 3,15,27,27,39 Will be 132 more than its 54th term

Answer 1

now first find 54th term
a54=a1+ (n-1)d
a54=3+(54-1)12 where a1=3 and d=15-3=12
a54= 3+636
a54=639

now add 132 in this term we get
771 .
now we have to find that which term will be 771 means we have to find n.
so 771=3+(n-1)12
771=3+12n-12
12n=780
n=780/12
n=65 so it will be 65th term.

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

Lets first calculate 54th of the given AP.

First term = a = 3

Common difference = d =15 - 3 = 12

Using formula an=a+(n−1)d,  to find nth term of arithmetic progression, we get

a54=a+(54−1)d

a54=3+53(12)=3+636=639

We want to find which term is 132 more than its 54th term. Lets suppose it is nth term which is 132 more than 54th term.

Therefore, we can say that

an=a54+132 

......{an=a+(n−1)d=3+(n−1)(12)}{a54=639}

⇒3+(n−1)12=639+132

⇒3+12n−12=771

⇒12n−9=771

⇒12n=780

⇒n=780/12=65

Therefore, 65th term is 132 more than its 54th term.

I hope, this will help you.

Thank you______❤

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